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C h a p t e r 1 | 1

Introduction to Electric Circuits

1.0 INTRODUCTION

This chapter is explaining about the basic principle of electric circuits and its connections.

The learning outcome for this chapter are the students should be able to explain clearly

basic electrical quantities, types of electrical circuits, electrical power, electrical energy

and solve related problems.

1.1 ELECTRIC

Electric is an energy which cannot see but can be felt and be used by human on today and

future. Electric energy can be created impact from action as friction, heat and

electromagnetic field

Electric energy can be change into other form of energy such as:

a) Light energy - lamp

b) Heat energy - Iron

c) Sound energy - Radio

d) Kinetic energy - Motor

There are two types of electric which is the static electric and dynamic electric.

a) Static electricity – A situation where no electron movement in certain direction.

b) Dynamic electricity – A situation where got electron movement in certain direction.

1.1.1 Electrical Quantities

i. Electromotive force (e.m.f)

Force or electric pressure that cause the flow of electrons or the flow of current

in given circuiy. The example the source that produces electric energy are

batteries and generator.

Symbol : E

Unit : Volt(V)

1 CHAPTER

INTRODUCTION TO

ELECTRIC CIRCUITS

C h a p t e r 1 | 2


ii. Electrical charge

There are two types of charge which is positive and negative charge. Electric

charge is measured in Coulomb.

Symbol : Q

Unit : Coulomb(C)

iii. Current

The movement of the electric charge cause by free electrons movement. It flows

from the positive terminal to the negative terminal.

Symbol : I

Unit : Ampiar (A)

iv. Voltage

The potential different between two points in a circuit.

Symbol : V

Unit : Volt(V)

v. Resistance

It is the property of material by which it oppose the flow of current through it.

Symbol : R

Unit : Ohm (Ω)

vi. Conductor

A material that allow electric current to flow easily. An example is copper and

iron.

vii.Insulator

A material that does not allow or prevent the electrical current flow in normal

condition. It has a lot of valence electrons but the valence electron are difficult to

be free from is parent atom. For example rubber, glass, air.

viii.Semiconductor

A material that has a conductance value between conductor and insulator. It has 4

valences electron and can be use to make electronic component. For examples

silicon and germanium.

xi.Resistivity

It is the characteristic of conductive material to opposition or decrease the current

flows in it.

Symbol : ρ (Rho)

Unit : Ohm meter ( )( mΩ

C h a p t e r 1 | 3


1.1.2 Resistance

The resistance of given material depends on the physical properties of the material.

There are 4 factor that influence the value of resistance:

i. Length of conductor, l

The length of conductor is proportional to the resistance. The longer the

length of the wire, the higher the resistance value.

l∝R

ii. Surface area , A

Area is inverse proportional to the resistance. As the resistance increase the

cross section area of a conductor will decreases.

AR

1∝

iii. Resistivity

Resistivity is proportional to the resistances. Higher the resistance, higher

the resistivity.

ρ∝R

iv. Conductor Temperature, T

The conductor temperature is proportional to the resistance. As the

conductor temperature increase the value of resistance also increase.

TR ∝

Mathematically, formula for the resistance of a wire of length l and the

cross section area A is as equation below:

(1.1)

Where: A = cross section area ( m2)

ρ = Resistivity ( mΩ )

l= Length (m)

R = Resistance ( Ω )

AR

lρ=

C h a p t e r 1 | 4


Example 1.1

Calculate the resistance of a 1.5 km length of aluminium wire. Given

diameter wire 10mm and the resistivity of aluminium is 0.025 µΩ.m.

Solution 1.1

Been given, mxd31010 −= , mx

3105.1=l , mx Ω= −610025.0ρ

Use equation, A

Rlρ

= ,

Where 2623

2 1054.78)2

1010()

2( mx

xdA

−−

=== ππ

Ω==∴−

−

477.01054.78

)105.1)(10025.0(6

36

x

xxR

1.2 ELECTRIC CIRCUIT

Electric circuit is a combination of conductor or cable which makes the current flow from

voltage sources to electrical components or load.There are two types of electric circuit:

i. Complete electric circuit

ii. Non Complete electric circuit

1.2.1 Complete electric circuit

It is also called basic circuit or simple circuit as shown in Figure 1.1. It is closed

end connection that can make current go through completely which the current

flow from source and back flow to sources again. The circuits must have voltage

supply (V), electric current (I) and resistance (R).

I

V R

Figure 1.1 : Complete Electric Circuit

1.2.2 Non Complete electric circuit

It is a circuit without one of three component either voltage sources or load

resistance. The current flow will never happen with perfect in non complete

circuit. There are two types of non complete circuit: Open circuit and Short

circuit

C h a p t e r 1 | 5


i. Open circuit

Is the circuit without the load, so there will be no current flow occur.

Value of resistant in this circuit is a higher. Figure 1.2 show a open

circuit.

V Pull out

load (R)

Figure 1.2 : Open Circuit

ii. Short circuit

The connection at the load will short with a conductor which no

resistance value as shown in Figure 1.3. The current which go through is

bigger. Normally if short circuit occur, the fuse will burnt.

I

V R Short with

a cable

Figure 1.3 : Short circuit

1.3 OHM’S LAW

Ohm’s law can be define as the current flowing through the electrical circuit is directly

proportional to the potential difference across the circuit and inversely proportional to

resistance of the circuit. If the value of resistance is constant and value of voltage

increase so the value of current can be increase. Mathematically the equation for Ohm’s

law is as equation 1.2 below.

(1.2)

where:

I = Current (A)

V = Voltage (V)

R = Resistance ( Ω )

IRV =

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The relationship between current and voltage is as shown using the graph at Figure 1.4.

This is the situation for constant value of resistance and temperature.

V (volt)

R (constant)

I (Ampere)

Figure 1.4 : Graph Voltage (V) vurses Current (I) for constant resistance

For the non constant or changing value of resistance, the relationship between voltage

and current are non linear as graph shown in Figure 1.5 below.

V

I Figure 1.5 : Graph Voltage (V) vurses Current (I) for non constant resistance

Example 1.2

Calculate the current value if the resistance is 10Ω and the supply voltage is 15V. Then,

calculates the new current value if the resistance has been change to 10 kΩ.

Solution 1.2

Been given , V= 15V

i) R = 10 Ω ,

Base on Ohm’s Law, V= IR

AR

VI 5.1

10

15===∴

ii) R = 10k Ω ,

mAxxR

VI 5.1105.1

1010

15 3

3====∴ −

C h a p t e r 1 | 7


1.4 ELECTRIC POWER

Electric power is a job can be done in one time unit. Resistor dissipate energy in the

form of heat. So power absorbed by the resistor is given by equation 1.3.

Symbol : P

Unit : Watt (W)

(1.3)

By using Ohm’s law, IRV = or R

VI = , It can derive new equation for power as in

equation 1.4.

RIP2=

(1.4)

R

VP

2

=

Where, P = Power (W),

I = Current (A)

R = Resistance (Ω) and

V = Voltage (V)

1.4.1 Wattmeter

Wattmeter is use to measure the value of the power that has been use. There are

two coils in wattmeter. The coil connected parallel to the load is voltage coil and

series with the load is current coil. The symbol for wattmeter is as shown in

Figure 1.6(a) and 1.6(b) is the internal wattmeter connection. Figure 1.7 is the

electric circuit connection using the wattmeter.

(a)Meter Symbol (b)Internal Connection

Figure 1.6 : Wattmeter

W

IVP =

C h a p t e r 1 | 8


Current coil

Voltage coil

Load (R)

Figure 1.7 : Electric Circuit Connection using Wattmeter

1.5 ELECTRIC ENERGY

Electric energy is a product of power and time. The symbol for electric energy is T or E.

Mathematically electric energy is expressed as equation 1.5.

1.5

where, T = electric energy (kWh)

P = power (W)

t = time (s)

V= voltage (V)

I = current (A)

R= resistance ( Ω )

Meter kilowatt-hour be used to measure total of electrical energy which be used by user.

The symbol metre 1.8. The electric power can be converted to horse power where 1

horse power = 746 watt

VS

tR

VT

RtIT

VItT

PtT

2

2

=

=

=

=

C h a p t e r 1 | 9


Figure 1.8 : Symbol Kilowatt Hour Metre

The unit for electric energy is Kilowatt hour (kWh) or Joule (J). When the current flow,

electron in the conductor will repell each other and it will produce heat and thus causing

the cabel that is used heating up.

Work is the energy absorbed to supply load 1 kW for 1 hour. Watt is the power used

when 1A current flows between 2 point that have 1 volt potential. Units for work is

Joule. This is equal to the energy produced to 1 Coulomb charge flows by 1 ohm

resistance. Total energy used to flow 1A current for 1 second by 1 ohm resistant is

called 1 Joule. It is can be called as 1 watt second, that is 1 watt power used for 1

second. In mathematical equation it can be shown as equation 1.6.

1.6

Example 1.3

A toaster taking 5A current from 240 V supply for 15 minutes. Calculate ,

i. Power used

ii. Energy absorbed in kJ

Solution 1.3

Given: I = 5 A , V = 240V dan t = 15 x 60 = 900s

i. WIVP 1200)240)(5( === .

ii. 1080000)900)(1200( === PtT W = 1080kWj = 1080 kJ

1.6 RESISTOR CIRCUIT ANALYSIS

Resistor can be connected in three different ways which are series, parallel and

combination of series and parallel.

1 Joule = 1 Watt second

work (J) = power (W) x time (s)

kWh

C h a p t e r 1 | 10


1.6.1 Series Circuit

Series circuit is refer to the connection of the resistor in the circuit. The resistors is

connected from end to end as in Figure 1.9. Series analysis are going to determine

total resistance, circuit current and total voltage.

IT

VT

Figure 1.9: Series Resistors

Total resistance, RT is the sum of all resistor which exist in the circuit. Equation

1.7 use to calculate total resistance.

(1.7)

Current through every resistor is equal to the total current, IT as show in equation

1.8:

(1.8)

Total voltage, VT is the sum of all voltage drops on every resistor as shown in

equation 1.9 below.

(1.9)

Voltage drop is the reduction of the voltage supply in every resistor. It can be

calculate using Ohm’s law and Voltage divider law. Voltage drop calculation

using ohm’s law are as Equation 1.10 below.

nTIIIII ==================== ......221

nTVVVVV ++++++++++++++++==== ......321

R1 R3

Rn

R2

+ V1 - + V2 - + V3 -

+

Vn

-

nTRRRRR ++++++++++++++++==== .....321

C h a p t e r 1 | 11


nTn

T

T

T

RIV

RIV

RIV

RIV

====

====

====

====

33

22

11

Meanwhile, to calculate voltage value across every resistance in series circuit

using voltage divider is as shown in equation 1.11 for three resistor connected in

series and equation1.12 is for two resistor connected in series.

TV

RRR

RV )(

321

1

1++++++++

====

TV

RRR

RV )(

321

2

2++++++++

==== (2.

TV

RRR

RV )(

321

3

3++++++++

====

TV

RR

RV )(

21

1

1++++

====

TV

RR

RV )(

21

2

2++++

====

Example 1.4 :

Refering to the circuit below determine :

R1 = 15ΩΩΩΩ

V = 120 V R2 =10ΩΩΩΩ

(1.10)

(1.11)

(1.12)

C h a p t e r 1 | 12


i). Total resistance, RT

ii). Current in the circuit, IT

iii). The voltage drop across each resistor.

Solution 1.4:


RT = R1 + R2 = (15 + 10) = 25 ΩΩΩΩ

ii). Current in the circuit, IT

IT = T

R

V =

25

120 = 4.8 A

iii). The voltage drop across each resistor.

VR1 = ITR1 = (4.8)(15) = 72 V

VR2 = ITR2 = (4.8)(10) = 48 V

1.6.2 Parallel Circuit

The parallel circuit is a connection of resistor which is against between each other.

The resistors connected in parallel is shown in Figure 1.10. Parallel analysis are

also going to determine total resistance, circuit current and total voltage.

IT I1 I2 I3 + + +

VT R1 V1 R2 V2 R3 V3

- - -

Figure 1.10: Parallel Resistor

The total parallel resistance can be calculate by using equation 1.13:

321

1111

RRRRT

++++++++====

Or

313221

321

RRRRRR

RRRR

T

++++++++====

(1.13)

C h a p t e r 1 | 13


The voltage across each parallel resistor is equal to the source voltage, VT as

shown in equation 1.14;

(1.14)

Total current, IT for parallel circuit is equal to the sumation of all current from

each branch. This is shown in equation 1.15

The value of the current for each branch also can be determine by using the

current divider’s law as equation 1.16.

IT I1 I2 + +

VT R1 V1 R2 V2

- -

Figure 1.11: Parallel Circuit using 2 Resistors

TI

RR

RI )(

21

2

1++++

====

and

T

IRR

RI )(

21

1

2++++

====

Example 1.5:

Refering to the circuit below, calculate :

IT I1 I2

R1 = 2ΩΩΩΩ R2 = 4ΩΩΩΩ

V = 240V

nTVVVVV ==================== ......221

(1.15)

(1.16)

nTIIIII ++++++++++++++++==== .....321

C h a p t e r 1 | 14



ii). Total current, IT

iii). Current I1 and I2

Solution 1.5:


RT = 21

21

RR

RR

+ =

42

)4)(2(

+ = 1.333 ΩΩΩΩ

ii). Total current, IT

T

T

T

R

VI ==== =

333.1

240 = 180 A

iii). Current I1 and I2 ,

AR

VI 120

2

240

1

1===

AR

VI 60

4

240

2

2===

1.6.3 Combination Circuit

Most of electric circuits are the combination of series and parallel circuit. Both

formula of series and parallel circuit will be used to determine the value of current,

voltage and total resistance. Figure 1.12 is the example for combination circuit.

R2 I2

R1 R3 I3

IT

V Figure 1.12 : Combination Series and Parallel Resistors

C h a p t e r 1 | 15


Example 1.6 :

By referring to Figure 1.12, a 120V source is connected across resistors. If R1 =

10Ω, R2 = 20Ω , R3 = 15Ω. Calculate

a). Total resistance, RT

b). Total current, IT

c). Current I2 and I3

Solution 1.6:

a). Total resistance, RT

32

3223

RR

RRR

+= = =

+ 1520

)15)(20( 8.57Ω

RT = R23 + R1 = 8.57 + 10 = 18.57ΩΩΩΩ

b). Total current, IT

IT = T

R

V =

57.18

120 = 6.46 A

c). Current

I2 = 46.6)1520

15()(

32

3

++++====

++++T

IRR

R = 2.79 A

∴ I3 = IT – I2 = (6.46 – 2.79) = 3.67 A

1.7 KIRCHOFF’S LAW

Kirchoff’s law is used to solve more difficult electric circuit, for example the circuit

which having more than one power supply. There are two types of Kirchoff law:

a) Kirchoff’s current law

b) Kirchoff’s voltage law

1.7.1 Kirchoff’s Current Law

Kirchoff’s current law is also known as first order of Kirchoff’s law. Kirchoff’s

current law stated that the algebraic sum of all the currents entering and leaving

a node is equal. Therefore, the sum of the current into a node (total current in) is

equal to the sum of the currents out of the node (total current out) as shown in

Figure 1.13.

i1 i2

i3

Figure 1.13: Flow of Current going In and Out the Node

C h a p t e r 1 | 16


In mathematic expression its can be stated as equation 1.17.

321 III += (1.17)

1.7.2 Kirchoff’s Voltage Law

Kirchoff’s voltage law also known as the second order of Kirchoff’s law. This

law stated that the sum of the voltage drop and voltage source around a closed

path is equal to zero.

+ V1 -

+

VT V2

-

+ V3 -

Figure 1.14: Close Path Voltage

This can be stated in mathematic expression like equation 1.18

321 VVVVT ++= (1.18)

Example 1.7:

R1 = 1Ω R2 = 6Ω R3 = 2Ω

5V 10V

By using Kirchoff’s current law, find the current thought each branch in the

circuit above.

C h a p t e r 1 | 17


Solution 1.7:

I1 A I2

I3

R1 = 1Ω R2 = 6Ω R3 = 2Ω

I II

5V 10V

Kirchoff’s Current Law:

I2 = - (I3 + I1 ) ……………………(1)

Kirchoff’s Voltage Law:

Loop I :

56

010)(6)1(5

31

31

−=−

=+−+−

II

II (2)

Loop II :

0)(2)6(10 23 =−+− II (3)

replace (1) into (3) :

1082

0)(2)(610

31

133

=+

=−−−+−

II

III (4)

Solve equations (2) and (4) by using Cramer rules.

i) Make the matrix equation from equation (2) and (4)

−=

−

10

5

82

61

3

1

I

I

ii) Find the value of determination, perhaps

−=

82

61D

20)2)(6()8)(1(82

61=−−=

−=∴ D

iii). Find the value of determination for each currents,

20)10)(6()8)(5(810

651 =−−−=

−−=I

C h a p t e r 1 | 18


AD

II 1

20

201

1 ===∴

AD

II

I

5.120

30

301020102

51

3

3

3

===∴

=+=−

=

From equation (1);

AIII 5.3)15.1()( 132 −=+−=+−=

Negative value (-ve) in the current 2I shows the actual current direction is

leading to resistance 2R .

REFERENCES

Bakshi, A.V. and Bakshi, U.A., 2009, “Circuit Theory 1st Edition”, Technical Publications

Pune, India

Bakshi, A.V. and Bakshi, U.A., 2008, “Circuit Analysis”, Technical Publications Pune, India

PROBLEMS

1. Calculate the resistance of the aluminium with 1.5 km length, 10 mm diameter and

0.025 µΩm resistivity.

(R=0.477Ω)

2. Calculate the resistance of the aluminium bar with 10m length, cross section area 8cm x

1 cm and 0.0269 µΩm resistivity.

(R=3.36 x 10-4Ω)

3. The cuprum with 2500cm and 1.75 µΩmm resistivity. Calculate the diameter of the

conductor when the resistor is 3.5 kΩ.

(d=1.26 x 10-5

m)

4. The heating element with 150 Ω resistance, 250 cm length and 0.7 mm. Calculate the

resistivity.

(R=2.3 x 10-5Ω)

5. Calculate the resistance for 31m length copper, 1.5 mm diameter and 0.017 µΩm

resistivity.

(R=0.298Ω)

C h a p t e r 1 | 19


6. The resistivity of the silinder aluminium conductor is 280 µΩmm, 1 mm radius and 30

Ω resistance. Calculate the length.

(ℓ=3.36 x 10-3

m)

7. Calculate the resistance of the conductor with 1.5 m length, 1.6 m2 cross section area

and 16.3 µΩm resistivity.

(R=15.28Ω)

8. Calculate the resistance of a zinc with 0.05 µΩm resistivity and 0.5m diameter.

(R=1.273 x 10-3Ω)

9. Calculate the current flowing through the aluminium wire with a length of 2 km and a

diameter of 20 mm if the 5V supply voltage. The resistivity of the wire is 0.28 µΩm.

10. Refer to the figure below, calculate the current flow in the conductor.

V=100V

ρ = 25.5µΩm d = 25 cm

100 km

(I=1.92A)

11. Calculate the current in the aluminium coil with 2 km length and 20 mm diameter when

the supply is 5V . The resistivity is 0.28 µΩ.m

(I=314.15A)

12. What is the current of a circuit that has 3 V and 0.5 ohm of resistance?

13. What is the voltage if current is 0.5 A [ampere] and resistance is 0.8 ohm?

14. What is the resistance of a circuit if voltage is 3 V and current is 2 A.

15. What is the current of a circuit if resistance is 3 ohm and voltage is 15V.

16. What is the voltage of a circuit if resistance is 7 ohm and current is 0.5 A.

17. Calculate the energy in Joule and Kilowatthour for:

a. A 60W lamp switched on for 8 hours

b. A 3kW kettle switched on for 5 minutes

18. Calculate the current flow in the circuit with 10Ω resistence and 15V voltage supply.

Then, calculate the current if the resistance is increasing to 10 kΩ.

19. Calculate the power losses when the current is 5mA through resistance 6kΩ.

20. Calculate the current flow by a Filament lamp with 240V and resistance 960Ω.

21. A cattle with resistance 40 Ω and current 2.4 A. Calculate the power.

(P=230.4w)

C h a p t e r 1 | 20


22. A toaster with 5A current and 240 V supply was on for 15 minutes. Calculate ,

i. Power used (1200w)

ii. Energy in kJ (1080kJ)

23. A Rice cooker with 3.45 kW power , 230 V voltage. Calculate:

i. Current (15A)

ii. Resistance (15.33Ω)

iii. Energy if the rice cooker is switch on for half an hour.(1.725kwh)

(6210kJ)

24. Calculate the amount of current (I) in a circuit, given below:

(4A)

25. Calculate the amount of resistance (R) in a circuit, given below

(9Ω)

26. Calculate the amount of voltage supplied by a battery in a circuit given below.

(14v)

C h a p t e r 1 | 21


27. Calculate:

I= 5 A

R= 10 Ω

V

i. Potential different, v

ii. Power, P

iii. Electrical energy if the circuit switch on for 2 hours

(50v, 250w, 0.5kwh)

28. Based on the circuit diagram below, calculate;

i. Total resistance.

ii. Total voltage, VT

iii. Voltage drop in the resistance R3 , the voltage divider law.

I = 1.5A R1 = 8 Ω R2 = 6 Ω

VT R3 = 4 Ω

(18 Ω, 27 V, 6 V)

29. Three (3) resistors connected in parallel to each value, Ω= 61R , Ω= 52R and

Ω= 203R and are supplied with a 100V. Calculate:

i. Total resistance

ii. Total current

iii. The voltage across each resistor

iv. Current through each resistor

(2.4 Ω, 41.7A, 100 V, 16.7A, 20A, 5A)

30. Define the First and Second order of Kirchoff’s Law

31. Based on figure below, find ;

i). Total Resistance.

ii). Voltage, R2 .

ii). Current R2 .and R3 .

iii). Total power and power dissipated in R1.

R3 = 8Ω

VT = 240V R1 = 2Ω R2 = 4Ω

(1.143Ω, 240v, 60A, 30A, 28.8kw, 50.4kw)

C h a p t e r 1 | 22


32. Based on figure below, the voltage across R1 = 72 V. Specify the following values :

i). The current flow each resistor R1, R2, R3 and R4

ii). The voltage across each resistor R2, , R3 and R4

iii). Supply voltage, Vs

R1 = 8Ω

IT

VS

R2 = 6Ω R3 = 3Ω

R4 = 4Ω

(9A, 3A, 6A, 18v, 36v, 126v)

33. Based on Figure below, calculate the current value of each branch and voltage drop in

each resistor using Kirchoff’s Law.

R2 = 4Ω R3 = 5Ω

12V 4V 6V

R1 = 1Ω (I2 = 1.5A, I3 = 0.5A, I1 = 2A, V1= 2v, V2= 6v, V3 = 2.5v)

34. Based on figure below, calculate the voltage supply.

i.

ii. I= 5 A

R= 10 Ω

V

(120A, 50A)

R = 20Ω 6 A

C h a p t e r 1 | 23


35. Based on figure below, calculate Total resistance, RT, Total current, IT, Voltage across

R1, VRI and Voltage across R2, VR2

i.

ii.

36. Based on figure below, calculate total resistance, RT

37. Based on figure below, calculate total resistance, RT referring to point AB.

i.

ii.

Vj

4Ω

1Ω

1Ω

1Ω

1Ω

1Ω

1Ω

4Ω 2Ω

R1 = 15Ω

Vj = 9v

R2= 25Ω

R3 = 5Ω

V1

V2 V3

R1 = 20Ω

Vj = 5v

R2= 10Ω

V1 V2

C h a p t e r 1 | 24


38. Based on the figure, calculate Total Resistance, RT, Total Current, IT, Current I1, Current

I2 and Total Power, PT

i.

ii.

39. Based on the figure, calculate Total Resistance, RT, Total Current, IT, Voltage across R1,

VRI , Voltage across R2, VR2 and Total Power, PT

40. Based on the figure, calculate Total Resistance, RT, Total Current, IT and Total Power, PT

R2 = 20Ω R4 = 12Ω

Vb

110 v

R1 = 10Ω R3 = 8Ω

2Ω 2Ω

6 v

Ij 2Ω

Vj = 10v

Ij

I1 I2

R1=5Ω R2=20Ω

Vj = 100v

Ij

I1 I2

R1=22Ω R2=28Ω

C h a p t e r 1 | 25


41. Based on the figure, calculate Total Resistance, RT, Total Current, IT, Voltage across

Resistor 30Ω, V30 , Voltage across resistor 20Ω, V20 and Voltage across resistor 50Ω, V50

42. Based on the figure, calculate Total Resistance, RT, Total Current, IT, Current I1, Current

I2 , Voltage across Resistor 8Ω, V1 and Voltage across resistor 4Ω, V2

43. By using the Ohm’s Law, Current Divider Law and Voltage Divider Law calculate Total

Current, IT, Current I1, Current I2 and Voltage drop at each resistors.

30Ω 50Ω

120 v

Ij 20Ω

8Ω 8Ω

Vb

12 v

4Ω

Ij I1 I2

R1 = 20Ω

R2 = 5Ω

R3 = 7Ω Ij

I1

I2

20 v

C h a p t e r 1 | 26


44. By using Kirchoff Law calculate Current I1, Current I2 and Current I3

i.

ii.

iii.

iv.

6Ω

12Ω 12 v

15v

3Ω I1 I2

I3

8Ω

5Ω

5 v 10 v

3Ω

I1 I2

I3

7 v

1

Ω

4 v

I1 I3 I2

1Ω

3 v

4Ω

2Ω 12 v

6v

4Ω I1 I2

I3

C h a p t e r 1 | 27


v.

180Ω

47Ω

150 Ω 4 v

47Ω

I1 I2

I3

2 v

100Ω

chapter - muhammad ahmad kamal's class · c h a p t e r 1 | 1 introduction to electric...

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